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Proof step 4 from proof lookup-put-rkvs-diff (conclusion)

⊦ r ≠ r0 → (st[((r -- sk) × val) ' + rkvs][r0 -- sk0]) = (st[rkvs][r0 -- sk0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ st[(rk × val) ' + rkvs] = (st[rk, val][rkvs])
Proof step 3:
⊦ r = r0, (st[r -- sk, val][rkvs][r0 -- sk0]) = (st[rkvs][r0 -- sk0])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (rkvs + rkvs0) + rkvs1 = rkvs + rkvs0 + rkvs1])]]